Page 4 - ln1

Solution:
In ΔABC, AB = AC (Given)
∴∠ACB = ∠ABC (Angles opposite to equal sides of a triangle are also equal)
In ΔACD, AC = AD
∴∠ADC = ∠ACD (Angles opposite to equal sides of a triangle are also equal)
In ΔBCD,∠ABC + ∠BCD + ∠ADC = 180º (Angle sum property of a triangle)
∠ACB + ∠ACB +∠ACD + ∠ACD = 180º
2(∠ACB + ∠ACD) = 180º
2(∠BCD) = 180º
∴∠BCD = 90º

4. In △ABC, D is the midpoint of BC. If DL ⊥AB and DM ⊥AC such that
DL = DM, prove that AB =AC.

Solution: It is given that D is the midpoint of BC
DL ⊥ AB and DM ⊥ AC such that DL = DM
Considering △ BLD and △ CMD as right angled triangle
o
So we can write it as∠ BLD = ∠ CMD = 90
We know that BD = CD and DL = DM
By RHS congruence criterion
△ BLD = △ CMD
∠ ABD = ∠ ACD (c. p. c. t)
Now, in △ ABC∠ ABD = ∠ ACD
We know that the sides opposite to equal angles are equal so we get
AB = AC

5.In △ABC, AB = AC and the bisectors ∠B and ∠C meet at a point O. Prove that
BO = CO and the ray AO is the bisector of ∠A.