Question

prove that △abd ≅ △acd in isosceles △abc, where ab = ac and ad is the altitude to bc. fill in the missing reason.

statement	reason
( angle adb = angle adc = 90^circ )	definition of altitude
( ad = ad )	reflexive property
( \triangle abd cong \triangle acd )	____

options:

a. asa congruence
b. aas congruence
c. sas congruenced. sss congruence

the midsegment of a triangle connects the midpoints of two sides and is always:

a. parallel to the third side and half its length.
b. equal to the third side.
c. perpendicular to the third side.
d. twice the length of the third side.

Explanation:

First Question (Triangle Congruence)

To determine the congruence criterion for \(\triangle ABD \cong \triangle ACD\):

• We know \(AB = AC\) (given), \(\angle ADB=\angle ADC = 90^\circ\) (altitude definition), and \(AD = AD\) (reflexive property).
• The sides \(AB\) and \(AC\) are the hypotenuses, \(AD\) is a leg common to both right triangles, and the included angle between \(AB\) (or \(AC\)) and \(AD\) is the right angle. So, we have two sides (\(AB = AC\), \(AD = AD\)) and the included angle (\(\angle ADB=\angle ADC = 90^\circ\)) equal. This matches the SAS (Side - Angle - Side) congruence criterion.

The midsegment theorem of a triangle states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half as long as the third side. Option a matches this theorem. Option b is incorrect as the midsegment is not equal to the third side. Option c is incorrect as the midsegment is not perpendicular to the third side (unless the triangle is isosceles or equilateral in a specific way, which is not general). Option d is incorrect as the midsegment is half the length of the third side, not twice.

Answer:

c. SAS Congruence

Second Question (Midsegment of a Triangle)