Q11 of 31 Page 232

AD and BE are the medians of triangle ABC. Let’s prove that triangle ACD = triangle BCE.

Given.


AD and Be are medians of triangle ABC


Formula used.


Median divides the triangle into 2 equal parts



As AD is median of BC of triangle ABC


triangle ADC = triangle ADB


Δ ADC + triangle ADB = triangle ABC


2 × triangle ADC = triangle ABC


triangle ADC = × triangle ABC


As BE is median of AC of triangle ABC


triangle BCE = triangle BAE


triangle BCE + triangle BAE = triangle ABC


2 × triangle BCE = triangle ABC


triangle BCE = × triangle ABC


By concluding both;


triangle ADC = triangle BCE


More from this chapter

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9

D is the midpoint of side BC of triangle ABC Parallelogram CDEF stands between side BC and parallel to BC through the point A. Let’s prove that triangle ABC = parallelogram CDEF.

 10

P is any a point on diagonal BD of parallelogram ABCD. Let’s prove that triangle APD = triangle CPD.

 12

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

 13

D is the midpoint of BC of triangle ABC and P is any a point on BC. Join P, A-Through the point D a straight line parallel to line segment PA meets AB at point Q. Let’s prove that,

(i) ΔADQ = triangle PDQ


(ii) ΔBPQ = ΔABC