# Find the size of angle PQA.In triangle ABC, angle BAC equals 50 degrees and angle ABC equals 55 degrees. P is a point on the side AB such that AP=PC and Q is a point on the side BC such that AQ...

Find the size of angle PQA.

In triangle ABC, angle BAC equals 50 degrees and angle ABC equals 55 degrees. P is a point on the side AB such that AP=PC and Q is a point on the side BC such that AQ bisects the angle BAC. Find the size of the angle PQA.

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You will want to draw the picture on a large sheet of paper and mark these as you go:

We are given that `m/_BAC=50,m/_ABC=55` so `m/_ACB=75`

`vec(AQ)` bisects `/_BAC` so `m/_BAQ=m/_AQC=25`

Since `bar(AP)cong bar(PC)` we have `m/_PAC=m/_PCA=50` from the isosceles triangle theorem.

Now `m/_PCQ=25` by subtraction. Then `m/_AQC=80` since the angle sum of `Delta AQC` is 180.

Let the intersection of `bar(AQ)"and"bar(PC)` be `R` .

We have `m/_QRC=m/_PRA=75` and `m/_PRQ=m/_ARC=105`

`Delta APR` ~`Delta CQR` by ` `AA~. Then `(AP)/(CQ)=(AR)/(CR)=(PR)/(QR)` since corresponding sides of similar triangles are in proportion.

Then `Delta PRQ` ~ `Delta ARC` by SAS~ since `m/_PRQ=m/_ARC=105` and `(AR)/(CR)=(PR)/(QR)==>(QR)/(RC)=(PR)/(AR)` (Exchange the extremes of the proportion)

That means that `m/_PQA=m/_PQR=m/_ACR=m/_ACP=50`

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**The measure of the required angle is 50**

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