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Question 9 In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

(i) Δ ABC ~ Δ AMP

(ii) $\dfrac{CA}{PA} = \dfrac{BC}{MP}$

Solution:

Given, ABC and AMP are two right triangles, right angled at B and M respectively.

(i) In ΔABC and ΔAMP, we have

$\angle CAB = \angle MAP$ (common angles)

$\angle ABC = \angle AMP = 90°$ (each 90°)

ΔABC ~ ΔAMP (AA similarity criterion)

(ii) As, ΔABC ~ ΔAMP (AA similarity criterion)

If two triangles are similar then the corresponding sides are always equal

Hence, $\dfrac{CA}{PA} = \dfrac{BC}{MP}$

Additional Questions

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:
(i)$\dfrac{CD}{GH} = \dfrac{AC}{FG}$
(ii) Δ DCB ~ Δ HGE
(iii) Δ DCA ~ Δ HGF

Answer

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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A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

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If AD and PM are medians of triangles ABC and PQR, respectively where Δ ABC ~ Δ PQR, prove that $\dfrac{AB}{PQ} = \dfrac{AD}{PM}$

Answer

State which pairs of triangles in Fig are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :

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In Fig 6.35, Δ ODC ~ Δ OBA, $\angle BOC = 125°$ and $\angle CDO = 70°$. Find $\angle DOC, \angle DCO$ and $\angle OAB$.

Answer

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that $\dfrac{OA}{OC} = \dfrac{OB}{OD}$

Answer
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