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In ∆PQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR. Complete the proof by filling in the boxes. - Geometry Mathematics 2

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प्रश्न

In ∆PQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR. 


Complete the proof by filling in the boxes.

In △PMQ, ray MX is bisector of ∠PMQ.

∴ `square/square = square/square` .......... (I) theorem of angle bisector.

In △PMR, ray MY is bisector of ∠PMQ.

∴ `square/square = square/square` .......... (II) theorem of angle bisector.

But `(MP)/(MQ) = (MP)/(MR)` .......... M is the midpoint QR, hence MQ = MR.

∴ `(PX)/(XQ) = (PY)/(YR)`

∴ XY || QR .......... converse of basic proportionality theorem.

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उत्तर

In △PMQ, ray MX is bisector of ∠PMQ.

∴ `bb(PX)/bb(XQ) = bb(MQ)/bb(MP)` .......... (I) theorem of angle bisector.

In △PMR, ray MY is bisector of ∠PMQ.

∴ `bb(PY)/bb(YR) = bb(MR)/bb(MP)` .......... (II) theorem of angle bisector.

But `(MP)/(MQ) = (MP)/(MR)` .......... M is the midpoint QR, hence MQ = MR.

∴ `(PX)/(XQ) = (PY)/(YR)`

∴ XY || QR .......... converse of basic proportionality theorem.

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अध्याय 1: Similarity - Problem Set 1 [पृष्ठ २८]

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बालभारती Mathematics 2 [English] Standard 10 Maharashtra State Board
अध्याय 1 Similarity
Problem Set 1 | Q 9 | पृष्ठ २८

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