Question

dont forget!
the definition of a rectangle
a quadrilateral where all four angles are equal is called a rectangle.
the property of a rectangle
the diagonals of a rectangle are equal in length.
1 in rectangle abcd on the right, point o is the intersection point of the diagonals. given co = 5 cm and ∠dac = 40°, answer the questions below.
(1) find the measure of ∠oab.
(2) find the measure of ∠ado.
(3) find the length of bo.
2 the following sentences describe a rectangle. fill the appropriate words in the blanks.
(1) in a rectangle, since the four are all equal, the two pairs of angles of the rectangle are equal respectively.
the quadrilateral in which the two pairs of opposite angles are equal respectively, confirms the property of a, so the rectangle is a parallelogram.
(2) in rectangle abcd on the right, from ∠abc = ∠dcb = 90° and bc is shared,
△abc ≅ △.
from this fact, ac = db is proved.
therefore, the length of the of a rectangle is shown to be equal.

Explanation:

Problem 1 (1): Find the measure of $\angle OAB$

Step 1: Recall rectangle angle property

In a rectangle, all angles are $90^\circ$, so $\angle DAB = 90^\circ$.

Step 2: Use given angle to find $\angle OAB$

Given $\angle DAC = 40^\circ$, and $\angle DAB=\angle DAC + \angle OAB = 90^\circ$. So $\angle OAB=90^\circ - 40^\circ = 50^\circ$.

Step 1: Recall rectangle diagonal property

In a rectangle, diagonals are equal and bisect each other, so $OA = OD$. Thus, $\triangle AOD$ is isosceles.

Step 2: Find $\angle ADO$

In $\triangle AOD$, $\angle DAC = 40^\circ$ (which is $\angle OAD$), so $\angle ADO=\angle OAD = 40^\circ$? Wait, no. Wait, $\angle DAB = 90^\circ$, $\angle OAB = 50^\circ$, and $OA = OB$ (diagonals bisect), but also in rectangle, $AD\parallel BC$, and diagonals bisect. Wait, correct approach: In rectangle, $\angle ADC = 90^\circ$, and $OA = OD$, so $\angle ODA=\angle OAD = 40^\circ$, then $\angle ADO = 40^\circ$? Wait, no, earlier for $\angle OAB$ we found $50^\circ$, and in $\triangle ABD$, $\angle OAB = 50^\circ$, $OA = OB$, so $\angle OBA = 50^\circ$, then $\angle AOB = 80^\circ$. But for $\angle ADO$: since $OA = OD$, $\angle OAD = 40^\circ$, so $\angle ADO=\angle OAD = 40^\circ$? Wait, no, $\angle DAC = 40^\circ$, which is $\angle OAD$, so in $\triangle AOD$, $OA = OD$, so base angles are equal. So $\angle ADO=\angle OAD = 40^\circ$? Wait, no, that can't be. Wait, $\angle DAB = 90^\circ$, $\angle DAC = 40^\circ$, so $\angle BAC = 50^\circ$. In rectangle, $AD\perp AB$, so $\angle ADB$: since diagonals are equal and bisect, $OA = OD$, so $\angle ODA=\angle OAD = 40^\circ$, so $\angle ADO = 40^\circ$? Wait, maybe I made a mistake. Wait, let's re - do:

In rectangle $ABCD$, $\angle DAB = 90^\circ$, $\angle DAC = 40^\circ$, so $\angle BAC = 50^\circ$. Diagonals $AC$ and $BD$ bisect each other, so $OA = OB = OC = OD$. So $OA = OD$, so $\triangle AOD$ is isosceles with $OA = OD$. Thus, $\angle ODA=\angle OAD = 40^\circ$, so $\angle ADO=\angle ODA = 40^\circ$? Wait, no, $\angle ADO$ is $\angle ODA$, so yes, $40^\circ$? Wait, but the first part's answer was $50^\circ$ for $\angle OAB$, which is correct because $\angle DAB = 90^\circ$, $\angle DAC = 40^\circ$, so $90 - 40 = 50$. Then for $\angle ADO$: in $\triangle AOD$, $OA = OD$, so $\angle OAD = 40^\circ$, so $\angle ADO = 40^\circ$. Wait, but maybe another way: $\angle ADC = 90^\circ$, $\angle DAC = 40^\circ$, so $\angle ACD = 50^\circ$. And $OD = OC$, so $\triangle ODC$ is isosceles, $\angle ODC=\angle OCD = 50^\circ$, then $\angle ADO=\angle ADC-\angle ODC = 90^\circ - 50^\circ = 40^\circ$. Yes, that matches. So $\angle ADO = 40^\circ$? Wait, no, the first part's answer for (1) was $50^\circ$, and maybe I messed up. Wait, no, the user's image shows (1) answer as $50^\circ$, so (2) let's see: $\angle ADO$: in rectangle, $AD\parallel BC$, $AB\perp AD$, so $\angle ADB$: since $OA = OD$, $\angle OAD = 40^\circ$, so $\angle ADO = 40^\circ$? Wait, no, maybe I made a mistake. Wait, $\angle DAB = 90^\circ$, $\angle OAB = 50^\circ$, so $\angle OAD = 40^\circ$, and $OA = OD$, so $\angle ADO=\angle OAD = 40^\circ$. So the measure of $\angle ADO$ is $40^\circ$? Wait, but let's check with triangle angles. In $\triangle AOD$, angles sum to $180^\circ$, so $\angle AOD = 180 - 40 - 40 = 100^\circ$. Then $\angle DOC = 80^\circ$, and since $OD = OC$, $\angle ODC=\angle OCD = 50^\circ$, which is consistent with $\angle ADC = 90^\circ$ (since $\angle ADO+\angle ODC = 40 + 50 = 90$). Yes, that works. So $\angle ADO = 40^\circ$. Wait, but the user's image for (1) has answer $50^\circ$, so (2) should be $40^\circ$? Wait, no, maybe I mixed up. Wait, the problem says $\angle DAC = 40^\circ$, which is $\angle OAD$, so $\angle ADO=\angle OAD = 40^\circ$. So the measure is $40^\circ$.

Step 1: Recall rectangle diagonal property

In a rectangle, diagonals are equal and bisect each other. Given $CO = 5$ cm, and diagonals bisect each other, so $BO = CO = 5$ cm (since $AC$ and $BD$ bisect at $O$, so $BO = OD$ and $AO = OC$, and $AC = BD$, so $BO = OC$).

Step 2: Conclude length of $BO$

Since $CO = 5$ cm, $BO = CO = 5$ cm.

Answer:

$50^\circ$

Problem 1 (2): Find the measure of $\angle ADO$