# 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

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

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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