What is the ncert class 9^{th} question of chapter triangles of exercise 7.1 question number 8(1). Please give me the simplest and easiest solution of this question , also give me the best solution of this question. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see Fig. 7.23). Show that: (i) ΔAMC ΔBMD .

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# In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see Fig. 7.23). Show that: (i) ΔAMC ΔBMD .Q.8(1)

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It is given that M is the mid-point of the line segment AB, C = 90°, and DM = CM

(i) Consider the triangles ΔAMC and ΔBMD:

AM = BM (Since M is the mid-point)

CM = DM (Given in the question)

CMA = DMB (They are vertically opposite angles)

So, by

SAS congruency criterion, ΔAMC ΔBMD.