Question

In the figure, P is the midpoint of AB and CD. a) Which angle has the same measure as that of angle APC ? b) Prove that AC = BD . c) Prove that BD is parallel to AC B A​

Answer 1

. In quadrilateral ABCD we have

AC = AD

and AB being the bisector of ∠A.

Now, in ΔABC and ΔABD,

AC = AD

[Given]

AB = AB

[Common]

∠CAB = ∠DAB [∴ AB bisects ∠CAD]

∴ Using SAS criteria, we have

ΔABC ≌ ΔABD.

∴ Corresponding parts of congruent triangles (c.p.c.t) are equal.

∴ BC = BD.

Answer 2

Answer:

In fig., O is the mid-point of AB and CD.

Prove that AC = BD and AC || BD. In the given figure, O is the middle point of both AB and CD.

In Δ’s AOC and BOD OA = OB OC = OD ∠AOC = ∠BOD (vertically opposite angles) ΔAOC = ΔBOD (by SAS congruence property) Hence AC = BD (by c.p.c.t) Also ∠OAC = ∠OBD (alt. angle) AC || BD Hence proved