In ΔPQR,
$\angle PQR = \angle PRQ$
∴ PQ = PR ………………………(i)
Given,
$ \dfrac{QR}{QS} = \dfrac{QT}{PR}$ Using equation (i), we get
$\dfrac{QR}{QS} = \dfrac{QT}{QP}$……………….(ii)
In ΔPQS and ΔTQR, by equation (ii),
$\dfrac{QR}{QS} = \dfrac{QT}{QP}$
$\angle Q = \angle Q$
∴ ΔPQS ~ ΔTQR [By SAS similarity criterion]
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S and T are points on sides PR and QR of Δ PQR such that $\angle P = \angle RTS$. Show that Δ RPQ ~ Δ RTS. |
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In Fig. 6.37, if Δ ABE ≅ Δ ACD, show that Δ ADE ~ Δ ABC.
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In Fig. 6.38, altitudes AD and CE of Δ ABC intersect each other at the point P. Show that:
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E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that Δ ABE ~ Δ CFB. |
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In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
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CD and GH are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that D and H lie on sides AB and FE of Δ ABC and Δ EFG respectively. If Δ ABC ~ Δ FEG, show that: |
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In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that Δ ABD ~ Δ ECF.
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Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of Δ PQR (see Fig. 6.41). Show that Δ ABC ~ Δ PQR.
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D is a point on the side BC of a triangle ABC such that $\angle ADC = \angle BAC$. Show that $CA^2$ = CB.CD. |
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Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that Δ ABC ~ Δ PQR. |
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