Question - 23 : - Prove that if the two arms of an angle are perpendicular to the two arms of another angle, then the angles are either equal or supplementary.

Answer - 23 : -

Given : In two angles ∠ABC and ∠DEF AB ⊥ DE and BC ⊥ EF

To prove: ∠ABC + ∠DEF = 180° or ∠ABC = ∠DEF

Construction : Produce the sides DE and EF of ∠DEF, to meet the sides of ∠ABC at H and G.

Proof: In figure (i) BGEH is a quadrilateral

∠BHE = 90° and ∠BGE = 90°

But sum of angles of a quadrilateral is 360°

∴ ∠HBG + ∠HEG = 360° – (90° + 90°)

= 360° – 180°= 180°

∴ ∠ABC and ∠DEF are supplementary

In figure (if) in quadrilateral BGEH,

∠BHE = 90° and ∠HEG = 90°

∴ ∠HBG + ∠HEG = 360° – (90° + 90°)

= 360°- 180° = 180° …(i)

But ∠HEF + ∠HEG = 180° …(ii) (Linear pair)

From (i) and (ii)

∴ ∠HEF = ∠HBG

⇒ ∠DEF = ∠ABC

Hence ∠ABC and ∠DEF are equal or supplementary

Question - 24 : - In the figure, lines AB and CD are parallel and P is any point as shown in the figure. Show that ∠ABP + ∠CDP = ∠DPB.

Answer - 24 : -

Given : In the figure, AB || CD

P is a point between AB and CD PD

and PB are joined

To prove : ∠APB + ∠CDP = ∠DPB

Construction : Through P, draw PQ || AB or CD

Proof: ∵ AB || PQ

∴ ∠ABP = BPQ …(i) (Alternate angles)

Similarly,

CD || PQ

∴ ∠CDP = ∠DPQ …(ii)

(Alternate angles)

Adding (i) and (ii)

∠ABP + ∠CDP = ∠BPQ + ∠DPQ

Hence ∠ABP + ∠CDP = ∠DPB

Question - 25 : -
In the figure, AB || CD and P is any point shown in the figure. Prove that:
∠ABP + ∠BPD + ∠CDP = 360°

Answer - 25 : -

Given : AB || CD and P is any point as shown in the figure

To prove : ∠ABP + ∠BPD + ∠CDP = 360°

Construction : Through P, draw PQ || AB and CD

Proof : ∵ AB || PQ

∴ ∠ABP+ ∠BPQ= 180° ……(i) (Sum of co-interior angles)

Similarly, CD || PQ

∴ ∠QPD + ∠CDP = 180° …(ii)

Adding (i) and (ii)

∠ABP + ∠BPQ + ∠QPD + ∠CDP

= 180°+ 180° = 360°

⇒ ∠ABP + ∠BPD + ∠CDP = 360°

Question - 26 : - In the figure, arms BA and BC of ∠ABC are respectively parallel to arms ED and EF of ∠DEF. Prove that ∠ABC = ∠DEF.

Answer - 26 : -

Given : In ∠ABC and ∠DEF. Their arms are parallel such that BA || ED and BC || EF

To prove : ∠ABC = ∠DEF

Construction : Produce BC to meet DE at G

Proof: AB || DE

∴ ∠ABC = ∠DGH…(i) (Corresponding angles)

BC or BH || EF

∴ ∠DGH = ∠DEF (ii) (Corresponding angles)

From (i) and (ii)

∠ABC = ∠DEF

Question - 27 : - In the figure, arms BA and BC of ∠ABC are respectively parallel to arms ED and EF of ∠DEF. Prove that ∠ABC + ∠DEF = 180°.

Answer - 27 : -

Given: In ∠ABC = ∠DEF

BA || ED and BC || EF

To prove: ∠ABC = ∠DEF = 180°

Construction : Produce BC to H intersecting ED at G

Proof: ∵ AB || ED

∴ ∠ABC = ∠EGH …(i) (Corresponding angles)

∵ BC or BH || EF

∠EGH || ∠DEF = 180° (Sum of co-interior angles)

⇒ ∠ABC + ∠DEF = 180° [From (i)]

Hence proved.

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