Q14 of 31 Page 232

In triangle ABC of which AB = AC; perpendiculars through the points B and C on sides AC and AB intersect sides AC and AB at the points E and F. Let’s prove that, FE || BC.

Given.


triangle ABC is isosceles triangle as AB = AC


BE and CF are perpendicular on AB and AC


Formula used.


AAS congruency rule = If 2 angles and one side of both triangles are equal then both triangle are congruent.


If 2 triangles are on same base And having equal area then they lies between 2 parallel lines.



In triangle AEB and triangle AFC


AB = AC ∵ Given


∠ AEB = ∠ AFC ∵ Both are perpendicular (90°)


∠ A = ∠ A ∵ common


triangle AEB ≅ triangle AFC [By AAS property]


If triangle AEB = triangle AFC


Then opening the triangles we get,


In triangle FEB + triangle AFE = triangle FEC + triangle AFE


On subtracting we get


triangle FEB = triangle FEC


In triangle FEB and triangle FEC


triangle FEB = triangle FEC


And they are on same base FE


Hence; they are between parallel lines


FE || BC


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A line parallel to BC of triangle ABC intersects sides AB and AC at the points P and Q respectively. CQ and BQ intersect each other at the point X. Let’s prove that.

(i) ΔBPQ = triangle CPQ


(ii) ΔBCP = triangle BCQ


(iii) ΔACP = triangle ABQ


(iv) ΔBXP = triangle CXQ

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(i) ΔADQ = triangle PDQ


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