# Congurency In Triangles

## Exercise 7.1

Question 1: In quadrilateral ABCD, AC = AD and AB bisects ∠A. Show that ΔACB ≅ ΔABD. AC=AD
∠CAB=∠DAB (AB is bisecting ∠CAD)
AB=AB (common side)
So, SAS axiom it is proved that:
ΔACB ≅ ΔABD

Question 2: ABCD is a quadrilateral in which AD=BC and ∠DAB=∠CAB

Prove that

1. ΔABD≅ ΔBAC
2. BD=AC
3. ∠ABD=∠BAC AB=AB (common side)
So, by SAS rule ΔABD∝ ΔBAC

Since ΔABD≅ ΔBAC
So, BD=AC
(Third corresponding sides of respective tri∠s)

In congruent triangles all corresponding angles are always equal.

Question 3: AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB. BC=AD (given)
∠CBO=∠DAO (right angle)
∠BOC=∠AOD (opposite angles)
So, by ASA rule
ΔBOC≅ ΔAOD
Or, BO=AO
And it is proved that CD bisects AB.

Question 4: l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that

ΔABC ≅ Δ CDA AB=CD (l and m are parallel)
AD=BC (AB and CD are parallel)
∠ABC=∠DCm (angles on the same side of transversal BC)
∠DCm=∠ADC
So, by SAS rule ΔABC≅ΔCDA

Question 5: Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of angle A. Show that:

1. ΔAPB ≅ ΔAQB
2. BP = BQ or B is equidistant from the arms of  angle A. AB=AB (common side)
∠PAB=∠QAB (AB is bisector of ∠QAP)
∠AQB=∠APB (right angle)
So, by ASA rule ΔAPB ≅ ΔAQB
And BQ=BP

Question 6: In the given figure, AC = AE, AB = AD and angle BAD = angle EAC. Show that BC = DE. AB=AD (given)
AC=AE (given)
Since ∠BAD=∠EAC
So, ∠BAD+∠DAC=∠EAC+∠DAC
Or, ∠BAC=∠DAE
So, by SAS rule ΔABC ≅ ΔADE
Or, BC=DE proved

Question 7: AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that angle BAD = angle ABE and angle EPA = angle DPB Show that

1. ΔDAP≅ ΔEBP
2. AD=BE ∠BAD=∠ABE (given)
∠EPA=∠DPB (given)
So, ∠EPA+∠EPD=∠DPB+∠EPD
Or, ∠DPA=∠EPB
AP=PB (P is midpoint of AB)
So, by ASA rule ΔDAP≅ ΔEBP
So, AD=BE

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. Show that:

1. ΔAMC ≅ ΔBMD
2. ∠DBC is a right angle
3. ΔDBC ≅ ΔACB
4. CM=1/2\AB BM=AM (M is midpoint)
DM=CM (given)
∠DMB=∠AMC (opposite angles)
So, ΔAMC≅ ΔBMD
Hence, DB=AC
∠DBA=∠BAC
So, DB||AC (alternate angless are equal)
So, ∠BDC=∠ACB= Right angle
(Internal angles are complementary in case of transverse of parallel lines)

In ΔDBC and ΔACB
DB=AC (proven earlier)
BC=BC (common side)
∠BDC=∠ACB (proven earlier)
So, ΔDBC≅ ΔACB
So, AB=DC
So, AM=BM=CM=DM
So, CM=1/2\AB