Math, asked by vikky721, 10 months ago

The diagonal AC of a parallelogram ABCD intersects DP at the point Q, where 'P' is any point on side AB. Prove that CQ × PQ = QA × QD

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Answered by Anonymous

131

★ Gɪᴠᴇɴ

▱ABCD is a parallelogram.
P is a point on AB
DP and AC intersect at Q.

☢ R.T.P

CQ . PQ = QA . QD

☛ Pʀᴏᴏғ

In ∆CQD, ∆AQP

∠QCD = ∠QAP

∠CQD = ∠AQP

∴ ∠QDC = ∠QPA

(∵ Angle sum property of triangles)

This, ∆CQD ∼ ∆ AQP by AAA similarity condition

∴ CQ/AQ = QD/QP = CD/AP

[∵ Ratio of corresponding sides of similar triangles are equal]

⇒ CQ/AQ = QD/QP

⇒ CQ . PQ = QA . QD [Q.E.D]

☞ Hence Proved ☜

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Answered by Anonymous

149

Step-by-step explanation:

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$\pink{\text{ABCD is a parallelogram}}$ .

$\purple{\text{P is a point on AB}}$

$\blue{\text{DP and AC intersect at Q.}}$

$\tt\red{\text{R.T.P}}$

$\tt\implies CQ . PQ = QA . QD$

$\huge{\blue{\fbox{\purple{\bigstar{\mathbf{\pink{proof}}}}}}}$

In ∆CQD, ∆AQP

∠QCD = ∠QAP

∠CQD = ∠AQP

∴ ∠QDC = ∠QPA. (∵ Angle sum property of triangles)

This,

∆CQD ∼ ∆ AQP ....................... AAA similarity condition

$\tt\implies\frac{CQ}{AQ}=\frac{QD}{QP}=\frac{CD}{AP}=$ ....................[∵ Ratio of corresponding sides of similar triangles are equal]

$\tt\implies \frac{CQ}{AQ}=\frac{QD}{QP}$

$\tt\implies CQ ×PQ = QA ×QD [Q.E.D]$

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#ʙᴇ ʙʀᴀɪɴʟʏ

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The diagonal AC of a parallelogram ABCD intersects DP at the point Q, where 'P' is any point on side AB. Prove that CQ × PQ = QA × QD​

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★ Gɪᴠᴇɴ

☢ R.T.P

☛ Pʀᴏᴏғ

☞ Hence Proved ☜

The diagonal AC of a parallelogram ABCD intersects DP at the point Q, where 'P' is any point on side AB. Prove that CQ × PQ = QA × QD