In the given figure ΔBCD is a square and ΔAPB is an equilateral tria

1.	In the given figure ΔBCD is a square and ΔAPB is an equilateral triangle. Prove that ΔAPD ≅ ΔBPC.[Hint: In ΔAPD and ΔBPC; \(\overline{AD} = \overline{BC}, \overline{AP} = \overline = {BP}\) and ∠PAD = ∠PBC = 90° – 60° = 30°]
Answer» Given that □ABCD is a square. ΔAPB is an equilateral triangle. Now in ΔAPD and ΔBPC AP = BP ( ∵ sides of an equilateral triangle) AD = BC (∵ sides of a square) ∠PAD = ∠PBC [ ∵ 90° – 60°] ∴ ΔAPD ≅ ΔBPC (by SAS congruence)

In the given figure ΔBCD is a square and ΔAPB is an equilateral triangle. Prove that ΔAPD ≅ ΔBPC.[Hint: In ΔAPD and ΔBPC; \(\overline{AD} = \overline{BC}, \overline{AP} = \overline = {BP}\) and ∠PAD = ∠PBC = 90° – 60° = 30°]

Answer»

Given that □ABCD is a square.

ΔAPB is an equilateral triangle.

Now in ΔAPD and ΔBPC

AP = BP ( ∵ sides of an equilateral triangle)

AD = BC (∵ sides of a square)

∠PAD = ∠PBC [ ∵ 90° – 60°]

∴ ΔAPD ≅ ΔBPC (by SAS congruence)