Answer: A. "Segment AD bisects angle CAB." is the right answer.
Step-by-step explanation:
Given : In ΔABC ,AC≅AB.
⇒∠ACB=∠CBA....(1) (∵ angles opposite to equal sides of a triangle are equal )
Now in ΔACD and ΔABD
AD=AD (common)....(2)
Here we need one more statement to prove the triangles congruent that is only statement (A) fits in it.
If AD bisects ∠CAB then ∠CAD=∠BAD..(3)
Now again Now in ΔACD and ΔABD
∠ACB=∠CBA [from (1)]
AD=AD [common]
∠CAD=∠BAD [from (3)]
So by ASA congruency criteria ΔADC≅ΔABD.
Answer:
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Step-by-step explanation:
we know, ∠BGA = 90
In right angled triangle BGA
AB = 12 cm (diameter of circle)
⇒ (AB)2 = (AG)2 + (BG)2
⇒ 122 = 82 + (BG)2
⇒ BG = 4√5
In right angled triangle BCG
Let CG be x cm.
⇒ (BC)2 = (CG)2 + (BG)2
⇒ (BC)2 = x2 + (4√5)2 ----(1)
As we know,
⇒ (BC)2 = CG × CA ----(2)
From equation (1) and equation (2)
⇒ x2 + (4√5)2 = x × (x + 8)
⇒ x2 + (4√5)2 = x2 + 8x
⇒ 8x = 80
⇒ x = 10 cm
From equation (2)
⇒ (BC)2 = CG × CA
⇒ (BC)2 = 10 × 18 = 180
⇒ BC = √180 = 6√5
6 × 5 = 30
Answer:
12
Step-by-step explanation:
you add 92 with the 4 and get 96 then you divide the 8x with the 96 and get 12
Hope it helps <333
Answer:
the last one is correct answer
