"Question 8 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 the given figure). Show that: (i) ΔAMC ≅ ΔBMD (ii) ∠DBC is a right angle. (iii) ΔDBC ≅ ΔACB (iv) CM = AB
Class 9 - Math - Triangles Page 120"
Answers
In Congruent Triangles corresponding parts are always equal and we write it in short CPCT i e, corresponding parts of Congruent Triangles.
It is necessary to write a correspondence of vertices correctly for writing the congruence of triangles in symbolic form.
Criteria for congruence of triangles:
There are 4 criteria for congruence of triangles.
SAS( side angle side):
Two Triangles are congruent if two sides and the included angle of a triangle are equal to the two sides and included angle of the the other triangle.
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Given:
In
right angled ∆ABC,
∠C = 90°,
M is the mid-point of AB i.e, AM=MB & DM = CM.
To Prove:
i) ΔAMC ≅ ΔBMD
ii) ∠DBC is a right angle.
Proof:
(i) In ΔAMC & ΔBMD,
AM = BM (M is the mid-point)
∠CMA = ∠DMB (Vertically opposite angles)
CM = DM (Given)
Hence, ΔAMC ≅ ΔBMD
( by SAS congruence rule)
ii) since, ΔAMC ≅ ΔBMD
AC=DB. (by CPCT)
∠ACM = ∠BDM (by CPCT)
Hence, AC || BD as alternate interior angles are equal.
Then,
∠ACB + ∠DBC = 180° (co-interiors angles)
⇒
90° + ∠B = 180°
⇒ ∠DBC = 90°
Hence, ∠DBC = 90°
(ii) In ΔDBC & ΔACB,
BC = CB (Common)
∠ACB = ∠DBC (Right angles)
DB = AC ( proved in part ii)
Hence, ΔDBC ≅ ΔACB (by
SAS congruence rule)
(iii) DC = AB (ΔDBC ≅ ΔACB)
⇒ DM + CM =AB
[CD=CM+DM]
⇒ CM + CM
= AB
[CM= DM
(given)]
⇒ 2CM = AB
Hence, CM=1/2AB
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AM=BM (GIVEN)
DM=CM (GIVEN)
AND angle DMB=CMA
By SAS CONGRUENCY RULE, triangle AMC congruent toBMD
(ii):since AMC CONGRUENT TO DMB, ANGLE B=ANGLE C (cpct)
(iii):in triangle DBC AND ACB,
ANGLE B=ANGLE C (PROVED ABOVE)
AC =BD (since triangle amc congruent to BMD ,cpct)
BC=B.C. (COMMON)
BY SAS , TRIANGLE DBC CONGRUENT TO ACB