In △abc (m∠c=90°), the points d and e are the points where the angle bisectors of ∠a and ∠b intersect respectively sides bc and ac. point g∈ ab so that dg ⊥ ab and h∈ ab so that eh ⊥ ab. prove that △ceh and △cdg are isosceles. (preferably in statement reason)

SOLUTION-

Triangle ABC is a right angle at C. The angles bisectors of BAC and ABC meet BC and AC at D and E  respectively. The points H and G are the feet of the perpendiculars from E and D to AB.

As AD is an angle bisector,

⇒ ∠CAD = ∠BAD

⇒ ∠CAD  = ∠GAD            ( ∵ ∠BAD = ∠GAD)

The two right triangles CAD and GAD ( ∵ m∠G = 90° = m∠C ) share a hypotenuse and have equal acute angles,

⇒ ΔCAD = ΔGAD

⇒ CD = DG

∴ ΔCDG is an isosceles triangle. (Proved)

Likewise,

As BE is an angle bisector,

⇒ ∠CBE = ∠ABE

⇒ ∠CBE  = ∠HBE           ( ∵ ∠ABE = ∠HBE)

The two right triangles CBE and HBE ( ∵ m∠H = 90° = m∠C ) share a hypotenuse and have equal acute angles,

⇒ ΔCBE = ΔHBE

⇒ CE = EH

∴ ΔCEH is an isosceles triangle. (Proved)

the answer is d

step-by-step explanation:

y+3=-4(x-2)

y+3=-4x+8

  -3         -3  

y=-4x+5