Question

question 2
1 pts
select all the true statements about the figure given below?
options:

• $\triangle obc$ is equilateral triangle
• $be = cd$
• $ae = ad$
• $\triangle eob \cong \triangle doc$
• $ob = oc$

Explanation:

To determine the true statements, we analyze each option:

1. $\boldsymbol{\triangle OBC}$ is equilateral triangle:

There’s no info (like equal sides/angles) to confirm $\triangle OBC$ has all sides equal. So this is false.

1. $\boldsymbol{BE = CD}$:

We need to check if $BE$ and $CD$ are equal. From $\triangle EOB \cong \triangle DOC$ (if proven) or other congruences, but let’s check other options first.

1. $\boldsymbol{AE = AD}$:

In $\triangle AEO$ and $\triangle ADO$:

• $\angle AEO = \angle ADO = 90^\circ$ (right angles from the diagram).
• $AO$ is common.
• $\angle EAO = \angle DAO$ (since $AO$ is an angle bisector, or from symmetry in the triangle).

By HL (Hypotenuse-Leg) or AAS, $\triangle AEO \cong \triangle ADO$, so $AE = AD$. This is true.

1. $\boldsymbol{\triangle EOB \cong \triangle DOC}$:

• $\angle OEB = \angle ODC = 90^\circ$ (right angles).
• $\angle EOB = \angle DOC$ (vertical angles).
• $OB = OC$? Wait, no—wait, from $\triangle ABC$, if it’s isoceles with $AB = AC$, then $OB = OC$ (since $AO$ is the median/angle bisector). Assuming $AB = AC$ (symmetry), $OB = OC$. Then by AAS, $\triangle EOB \cong \triangle DOC$. So this is true.

1. $\boldsymbol{OB = OC}$:

If $\triangle ABC$ is isoceles with $AB = AC$, and $AO$ is the angle bisector/median, then $OB = OC$ (by the Isosceles Triangle Theorem, as $AO$ is the axis of symmetry). So this is true.

Wait, the original selected option is $AE = AD$ (true), and we also confirm $BE = CD$ (from $\triangle EOB \cong \triangle DOC$, so $BE = CD$), $\triangle EOB \cong \triangle DOC$ (true), and $OB = OC$ (true).

Final True Statements:

• $AE = AD$ (selected, and proven true).
• $BE = CD$ (from $\triangle EOB \cong \triangle DOC$).
• $\triangle EOB \cong \triangle DOC$ (proven true).
• $OB = OC$ (proven true).

(Note: The initial selection of $AE = AD$ is correct, and additional true statements are $BE = CD$, $\triangle EOB \cong \triangle DOC$, and $OB = OC$.)

Answer:

To determine the true statements, we analyze each option:

1. $\boldsymbol{\triangle OBC}$ is equilateral triangle:

There’s no info (like equal sides/angles) to confirm $\triangle OBC$ has all sides equal. So this is false.

1. $\boldsymbol{BE = CD}$:

We need to check if $BE$ and $CD$ are equal. From $\triangle EOB \cong \triangle DOC$ (if proven) or other congruences, but let’s check other options first.

1. $\boldsymbol{AE = AD}$:

In $\triangle AEO$ and $\triangle ADO$:

• $\angle AEO = \angle ADO = 90^\circ$ (right angles from the diagram).
• $AO$ is common.
• $\angle EAO = \angle DAO$ (since $AO$ is an angle bisector, or from symmetry in the triangle).

By HL (Hypotenuse-Leg) or AAS, $\triangle AEO \cong \triangle ADO$, so $AE = AD$. This is true.

1. $\boldsymbol{\triangle EOB \cong \triangle DOC}$:

• $\angle OEB = \angle ODC = 90^\circ$ (right angles).
• $\angle EOB = \angle DOC$ (vertical angles).
• $OB = OC$? Wait, no—wait, from $\triangle ABC$, if it’s isoceles with $AB = AC$, then $OB = OC$ (since $AO$ is the median/angle bisector). Assuming $AB = AC$ (symmetry), $OB = OC$. Then by AAS, $\triangle EOB \cong \triangle DOC$. So this is true.

1. $\boldsymbol{OB = OC}$:

If $\triangle ABC$ is isoceles with $AB = AC$, and $AO$ is the angle bisector/median, then $OB = OC$ (by the Isosceles Triangle Theorem, as $AO$ is the axis of symmetry). So this is true.

Wait, the original selected option is $AE = AD$ (true), and we also confirm $BE = CD$ (from $\triangle EOB \cong \triangle DOC$, so $BE = CD$), $\triangle EOB \cong \triangle DOC$ (true), and $OB = OC$ (true).

Final True Statements:

• $AE = AD$ (selected, and proven true).
• $BE = CD$ (from $\triangle EOB \cong \triangle DOC$).
• $\triangle EOB \cong \triangle DOC$ (proven true).
• $OB = OC$ (proven true).

(Note: The initial selection of $AE = AD$ is correct, and additional true statements are $BE = CD$, $\triangle EOB \cong \triangle DOC$, and $OB = OC$.)