Exercise 7.3: Triangles — NCERT Solutions

Given: $AB = AC$, and $BO$, $CO$ bisect $\angle B$ and $\angle C$ respectively.

To prove: $OB = OC$.

Proof:

Since $AB = AC$, the angles opposite them are equal (isosceles triangle property): $\angle ABC = \angle ACB.$

Taking halves of equal angles: $\tfrac12\angle ABC = \tfrac12\angle ACB \implies \angle OBC = \angle OCB.$

In $\triangle OBC$, two base angles are equal, so the sides opposite them are equal: $OB = OC.$

Given: $BE \perp AC$, $CF \perp AB$, and $BE = CF$.

To prove: $AB = AC$.

Proof:

In $\triangle BEC$ and $\triangle CFB$:

1. $\angle BEC = \angle CFB = 90^\circ$ (altitudes)
2. $BC = CB$ (common hypotenuse)
3. $BE = CF$ (given)

By the RHS congruence rule, $\triangle BEC \cong \triangle CFB.$

By CPCT, $\angle BCE = \angle CBF,\quad\text{i.e.}\quad \angle ACB = \angle ABC.$

Since the base angles of $\triangle ABC$ are equal, the sides opposite them are equal: $AB = AC.$

Hence $\triangle ABC$ is isosceles.

Given: $AB = AC$ and $AD = AB$, so $AD = AC$. $D$ lies on ray $BA$ produced.

To prove: $\angle BCD = 90^\circ$.

Proof:

In $\triangle ABC$, since $AB = AC$, $\angle ACB = \angle ABC = x \ (\text{say}).$

In $\triangle ACD$, since $AC = AD$, $\angle ACD = \angle ADC = y \ (\text{say}).$

Now $\angle BCD = \angle ACB + \angle ACD = x + y$.

In $\triangle BCD$, the angle sum gives: $\angle DBC + \angle BCD + \angle BDC = 180^\circ.$

Since $\angle DBC = \angle ABC = x$ and $\angle BDC = \angle ADC = y$: $x + (x + y) + y = 180^\circ \implies 2(x + y) = 180^\circ.$

Therefore $x + y = 90^\circ \implies \angle BCD = 90^\circ.$