Question

7. In the adjacent figure ABCD is a square and AAPB is an
equilateral triangle. Prove that AAPD: ABPC.
(Hint: In AAPD and ABPC, AD = BC, AP = BP and
ZPAD = Z PBC = 90° -60° = 30°]​

Answer 1

Answer:

∆ APD and ∆BPC

AD = BC (sides of square)

anglePAD = anglePBC = 30° (given)

AP = BP (sides of equilateral triangle)

SO, ∆APD is congurent to ∆BPC

Step-by-step explanation:

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