Question 1: In quadrilateral ABCD, `AC = AD` and AB bisects ∠A. Show that ΔACB ≅ ΔABD.

**Answer:** in ΔACB and 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

- `ΔABD≅ ΔBAC`
- `BD=AC`
- `∠ABD=∠BAC`

**Answer:** In ΔABD and ΔBAC

AD=BC

AB=AB (common side)

∠BAD=∠ABC

So, by SAS rule ΔABD∝ ΔBAC

Since ΔABD≅ ΔBAC

So, BD=AC

(Third corresponding sides of respective tri∠s)

In congruent tri∠s all corresponding ∠s are always equal.

So, ∠BAD=∠ABC proved

Question 3: AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.

**Answer:** In ΔBOC and ΔAOD

`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

**Answer:** In ΔABC and ΔCDA

`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:

- ΔAPB ≅ ΔAQB
- BP = BQ or B is equidistant from the arms of angle A.

**Answer:** In ΔAPB and ΔAQB

`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.

**Answer:** In ΔABC and ΔADE

`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

- `ΔDAP≅ ΔEBP`
- `AD=BE`

**Answer:** In ΔDAP and ΔEBP

`∠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:

- `ΔAMC ≅ ΔBMD`
- `∠DBC` is a right angle
- `ΔDBC ≅ ΔACB`
- `CM=1/2\AB`

**Answer:** In ΔAMC and ΔBMD

`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`

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