Solve Q9 its urgent

Solve Q9 its urgent DE Hint: CD - 8. In the adjoining figure, AB = AC and D is mid-point oi BC. Use SSS rule Of congruency to show that (i) AABDEA.ACD (ii) AD is bisector of ZA (iii) AD is perpendicular to BC, 9. Two line segments AB and CD bisect each other at O. Prove that (i) ACE BD (iii) AD'JCB (ii) ZCAB-ZABD (it') AD-CB. a AASeon triangles are equal. In tyæ (The result f. triangle are triangle. 'k at the • SABC, AABt Ve eon! 10. In the adjoining figure, find the values of x and y. 2.2 Some more criteria for congruence of triangles raw two triangles ABC and I'QR such that BC QR = LB ZQ = 600 and -450. Make a trace-copy of AABC and place it over APQR. We observe that the two triangles ver each other exactly and so they are congruent. Repeat this activity with more pairs of angles satisfying these conditions. We serve that the equality of two angles and the duded side is sufficient for the congruence of o triangles. We record it as: ASA rule of congruency Two triangles are congruent if two angles and the included side of one triangle are equal to two side of the nthcr triangle. Thus, is a p verti( (bec; angl Ins ay n/ 450

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In ?AOC and ? DOB
Oc = od
Ao = Bo
So ?s are congruent by sas Cong rule
Part a and b are by cpct
(C) angle cao is equal to angle obd which are alternate interior angles so the lines are ll.
Part d. Do your self take some stress about study
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Please find this answer

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IN TRIANGLE ABC D IS THE MIDPOINT OF BC SO JOIN D TO NOW AB=AC[GIVEN]
BD=DC[D IS THE MIDPOINT OF BC]
AD =AD [COMMON]
SO TRIANGLE ABD IS CONGRUENT TO ACD
THUS BY CPCT 2  QUESTIONS CAN BE PROOFED
ADB+ADC=180[LINEAR PAIR]
BUT ADB =ADC
SO ADB +ADB =18 SO ADB =90
HENCE AD IS THE PERPENDICULUR BISECTTOR


















 
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