Q15 of 56 Page 127

The mid point of side BC of ΔABC is D. If from B a line is drawn bisecting AD such that cutting side AD at E if cuts AC at X. Then prove that


Let a point F on AC such that DF||BX.


By Converse of Mid Point Theorem, as D is mid point of BC, F is the mid point of AC.


⇒ CF = XF


In ΔCFD & ΔCXB,


|By Mid Point Theorem


⇒ BX = 2DF


In ΔAXE & ΔAFD


E is the mid point of AD


and EX | | DF


By Mid Point Theorem,


AX = XF





Hence, proved.


More from this chapter

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13

Prove that the triangles formed by joining the mid – points of the three sides of a triangle consecutively are similar to their original triangle.

 14

As shown in the figure of AB ⊥ BC, DC ⊥ BC and DE ⊥ AC then prove that ΔCED ~ ΔABC.

 1

Answer the following in True or False. Write the reason of your answer (if possible).

(i) The ratio of the corresponding sides of two similar triangles is 4 : 9. Then the ratio of the areas of these triangles is 4 : 9.


(ii) In two triangles respectively ΔABC and ΔDEF of then ΔABC ≅ ΔDEF.


(iii) The ratio of the areas of two similar triangles in proportional to the squares of their sides.


(iv) If ΔABC and ΔAXY are similar and the values of their areas are the same then XY and BC may be coincident sides.

 2

If ΔABC~ΔDEF and their areas are respectively 64 sq cm and 121 sq cm. If EF = 15.4 cm then find BC.