In Fig. 4.145, If AB⊥BC, DC⊥BC and DE⊥AC, prove that ΔCED~ ΔABC.
Attachments:
Answers
Answered by
2
To Prove ΔCED~ ΔABC
Step-by-step explanation:
Given :
AB ⊥BC
DC ⊥BC
DE ⊥ AC
Proof :
In ΔABC, ∠ABC = 90° , as AB⊥BC
∠DCB = 90°, as DC⊥BC
∠ABC +∠DCB = 90°+90° = 180 °
The sum of Co -interior angles is 180°
In ΔABC and ΔDEC ,
∠ABC = ∠DEC = 90°
Alternate interior angles are always equal
∴∠BAC = ∠DCE = 90°
By Angle - Angle Similarity ,we prove that ΔCED~ ΔABC.
To Learn More.....
1. In triangle abc, de is parallel to side a b such that a b =2dc. if area of triangle abed =56sq. cm units then find area of triangle abc
https://brainly.in/question/8656457
2. Find the area of the adjoining heptagon ABCDEFG Fig. ii) in which AC =6.5 cm , BM = 3.2 cm , CE = 9 cm, DN = 4.5 cm , AG = 10 cm, AP = 4.4 cm.
https://brainly.in/question/9271901
Attachments:
Answered by
4
Answer:
To Prove ΔCED~ ΔABC
Step-by-step explanation:
Given :
AB ⊥BC
DC ⊥BC
DE ⊥ AC
Proof :
In ΔABC, ∠ABC = 90° , as AB⊥BC
∠DCB = 90°, as DC⊥BC
∠ABC +∠DCB = 90°+90° = 180 °
The sum of Co -interior angles is 180°
In ΔABC and ΔDEC ,
∠ABC = ∠DEC = 90°
Alternate interior angles are always equal
∴∠BAC = ∠DCE = 90°
Step-by-step explanation:
Similar questions