Question

measures of some angles in figure are given prove that AP/PB=AQ/QC​

Answer 1

We know that APQ=ABC=60°

Also, PAQ=BAC (same angles)

Therefore by AA axiom, Triangle APQ is similar to triangle ABC

Therefore,

AP/PB=AQ/QC

(Similar triangles have their ratio of same side equal)

Answer 2

Given:

∠P = 60° and ∠B = 60°.

To prove:

AP/PB = AQ/QC​

Proof:

1) In the given triangle ABC

• ΔABC & ΔAPQ

• ∠P = ∠B = 60° (given)

• ∠A = ∠A ( common)

• ∠C = ∠Q (ultimately equal if the other two are equal)

• ΔABC ≈ ΔAPQ ( by AA similarity rule)

2) So by the similarity ration we get

• $\frac{AB}{AP } = \frac{AC}{AQ} =\frac{BC}{PQ}$

• $\frac{AB}{AP} -1=\frac{AC}{AQ}-1$  (Subtract 1 from both side)

• $\frac{AB-AP}{AP}=\frac{AC-AQ}{AQ}$  

• $\frac{PB}{AP}=\frac{QC}{AQ}$

• $\frac{AP}{PB}=\frac{AQ}{QC}$ (reciprocal)

Hence proved.