Question

In the adjoining figure, in ΔABC, A-P-B and A-Q-C Prove that A(ΔAPQ)/A(ΔABC)=AP×AQ/AB×AC.

Answer 1

Step-by-step explanation:

ANSWER

Given Δ ABP is an equilateral triangle.

Then, AP = BP (Same side of the triangle)

and AD = BC (Same side of square)

and, ∠ DAP = ∠ DAB - ∠ PAB = 90° - 60° = 30°

Similarly,

∠ BPC = ∠ ABC - ∠ ABP = 90° - 60° = 30°

∴ ∠ DAP = ∠ BPC

∴ Δ APD is congruent to Δ BPC ( SAS proved)

In Δ APD

AP = AD II as AP = AB (equilateral triangle)

We know that ∠ DAP = 30°

∴ ∠ APD = (180° - 30°)/2 (Δ APD is an isosceles triangle and ∠ APD is on of the base angles.)

= 150°/2 = 75°

= ∠ APD = 75°

Similarly, ∠ BPC = 75°

Therefore, ∠ DPC = 360° - (75°+75°+60°)

= ∠ DPC = 150°

Now in Δ PDC

PD = PC as Δ APD is congruent to Δ BPC

∴ Δ PDC is an isosceles triangle

And ∠ PDC = ∠ PCD = (180° - 150°)/2

or ∠ PDC = ∠ PCD = 15°

∠ DPC = 150°; ∠ PDC = 15° and ∠ PCD = 15°

Answer 2

Step-by-step explanation:

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Answer♡

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