Answer:
Δ MLP ~ Δ HKG
By S-A-S similarity Postulate.
Step-by-step explanation:
Given:
In Δ MLP
MP = 20 ,
LP = 15
∠ P = 42°
In Δ HKG
HG = 16 ,
KG = 12
∠G = 42°
To Prove:
Δ MLP ~Δ HKG
Proof:
In Δ MLP and Δ HKG
[tex]\dfrac{MP}{HG}= \dfrac{20}{16}=\dfrac{5}{4}[/tex] ..........( 1 )
[tex]\dfrac{LP}{KG}= \dfrac{15}{12}=\dfrac{5}{4}[/tex] ..............( 2 )
∴ [tex]\dfrac{MP}{HG}=\dfrac{LP}{KG}[/tex] ...From 1 and 2
∠P ≅ ∠ G = 42° ...............Given
∴ Δ MLP ~Δ HKG ......{ By S-A-S similarity test}............Proved
