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ΔABC is an isosceles triangle with AB = AC, D is the midpoint of BC and
AD is joined. Prove that ΔABD ≅ ΔACD.
I kinda have a problem with triangles and congruency....little help please ?


Sagot :

Step-by-step explanation:

Okay, so the triangle would look something like this:

see the image attached

Since AB=AC, which is given; <ABD = <ACD, because they are the angles opposite from the congruent sides of the isosceles triangle; and <ADB=<ADC, as AD is the bisector*; ΔABD ≅ ΔACD using SAA/AAS (side-angle-angle) congruencey.

*If the bisector of an angle in a triangle meets the opposite side at its midpoint, then the triangle is isosceles, and vice versa

(I haven't done Geometry in a while so I hope that's correct :) )

View image doudoupeihe
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