# In a triangle ABC, P and Q are points on the sides AB and CB respectively such that (angle)APC = (angle)AQC, AP = 70cm, PB = 30cm, AQ = 120cm and CP = 150cm. Find the length of CQ.

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Given `Delta ABC` and point `P` on `bar(AB)` such that `AP=70,BP=30` and `CP=150` , point `Q` on `bar(BC)` such that `/_APC cong /_AQC` and `AQ=120` :Find `CQ` .

Let `X` be the intersection of `bar(CP),bar(AQ)` . Then `Delta APX`~`Delta CQX` by AA~ (Given two congruent angles and a pair of vertical angles). Then `/_PAQ cong /_QCP` .

Now `Delta BAQ` ~`Delta BCP` by AA~ (Angle B by the reflexive property).

This implies that `(BA)/(BC)=(BQ)/(BP)=(AQ)/(CP)` . Using the last proportion and substituting known values we get:

`(BQ)/(30)=(120)/(150)==>BQ=24`

Now using `(BA)/(BC)=(AQ)/(CP)` and plugging in known values we get:

`100/(BC)=(120)/(150)==>BC=125`

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`BC=BQ+CQ==>125=24+CQ==>CQ=101`

**The length of `bar(CQ)` is 101cm**

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